Graphs of functions of the form 2 at AX C. Lesson "Function Y \u003d AX2, its schedule and properties

The study of the properties of functions and their graphs occupies a significant place in both school mathematics and subsequent courses. And not only in courses of mathematical and functional analysis, and not only in other sections higher MathematicsBut in most narrow professional items. For example, in the economy - the functions of utility, costs, demand functions, supply and consumption ..., in radio engineering - control functions and response functions, in statistics - distribution functions ... To facilitate further study of special functions, you need to learn to freely operate the elementary graphs functions. To do this, after studying the next table, we recommend passing the "Conformation of Function Graphs" link.

In the school course of mathematics are studied the following
elementary functions.

Function name	Formula function	Schedule function	Graphic name	Comment
Linear	y \u003d kx.		Straight	The most simple private case of linear dependence is direct proportionality. y \u003d kx.where k. ≠ 0 - proportionality coefficient. In the picture, an example for k. \u003d 1, i.e. In fact, the given graph illustrates a functional dependence that specifies the equality of the value of the function value of the argument.
Linear	y. = kX. + b.		Straight	General linear dependence: coefficients k. and b. - Any valid numbers. Here k. = 0.5, b. = -1.
Quadratic	y \u003d X. 2		Parabola	The simplest case of quadratic dependence is a symmetric parabola with a vertex at the beginning of the coordinates.
Quadratic	y \u003d AX. 2 + bX. + c.		Parabola	General case of quadratic dependence: coefficient a. - an arbitrary valid number is not zero ( a. belongs r, a. ≠ 0), b., c. - Any valid numbers.
Power	y \u003d X. 3		Cubic Parabola	The easiest case for an odd degree. Cases with coefficients are studied in the "Motion of Function Graphs" section.
Power	y \u003d X. 1/2		Schedule function y. = √x.	The easiest case for fractional degree ( x. 1/2 = √x.). Cases with coefficients are studied in the "Motion of Function Graphs" section.
Power	y \u003d k / x		Hyperbola	The easiest case for a short degree ( 1 / x \u003d x -1) - back-proportional dependence. Here k. = 1.
Indicative	y. = e X.		Exhibitor	An exponential dependence is called an indicative function for the foundation. e. - irrational number of approximately equal to 2,7182818284590 ...
Indicative	y \u003d a x		Graph indicative function	a. \u003e 0 I. a. a.. Here is an example for y \u003d 2 x (a. = 2 > 1).
Indicative	y \u003d a x		Graph indicative function	Exponential function Defined for a. \u003e 0 I. a. ≠ 1. Fun graphics significantly depend on the value of the parameter a.. Here is an example for y \u003d 0.5 x (a. = 1/2 < 1).
Logarithmic	y. \u003d LN. x.			Graph Logo Function for Base e. (Natural logarithm) is sometimes called logarithmics.
Logarithmic	y. \u003d Log. A X.		Schedule logarithmic function	Logarithms are defined for a. \u003e 0 I. a. ≠ 1. Fun graphics significantly depend on the value of the parameter a.. Here is an example for y. \u003d log 2. x. (a. = 2 > 1).
Logarithmic	y \u003d Log. A X.		Schedule logarithmic function	Logarithms are defined for a. \u003e 0 I. a. ≠ 1. Fun graphics significantly depend on the value of the parameter a.. Here is an example for y. \u003d log 0.5 x. (a. = 1/2 < 1).
Sinus	y. \u003d SIN x.		Sinusoid	Trigonometric function sinus. Cases with coefficients are studied in the "Motion of Function Graphs" section.
Cosine	y. \u003d COS. x.		Kosinusoid	Trigonometric cosine function. Cases with coefficients are studied in the "Motion of Function Graphs" section.
Tangent	y. \u003d TG. x.		TangentSoid	Trigonometric function Tangent. Cases with coefficients are studied in the "Motion of Function Graphs" section.
Cotangent	y. \u003d CTG. x.		Kothangensoid	Trigonometric Cotangen feature. Cases with coefficients are studied in the "Motion of Function Graphs" section.

Inverse trigonometric functions.

Function name	Formula function	Schedule function	Graphic name

An abstract of lesson on algebra for grade 8 secondary school

Lesson Topic: Function

The purpose of the lesson:

Educational: Determine the concept of the quadratic function of the form (compare graphs of functions and), show the formula for finding the coordinates of the pearabera vertex (to teach it to apply this formula in practice); To form the ability to determine the properties of a quadratic function according to the graph (finding the axis of symmetry, the coordinates of the pearabol vertex, the coordinates of the intersection of the graph with the coordinate axes).

Developing: The development of mathematical speech, the ability is correct, consistently and rationally express their thoughts; Development of the skill of the correct recording of mathematical text using symbols and designations; development of analytical thinking; The development of cognitive activity of students through the ability to analyze, systematize and summarize the material.

Educational: upbringing independence, ability to listen to others, the formation of accuracy and attention in writing mathematical speech.

Type of lesson: Studying a new material.

Teaching methods:

generalized-reproductive, inductively heuristic.

Requirements for knowledge and skills of students

know what is a quadratic function of the species, the formula for finding the coordinates of the pearabol vertex; To be able to find the coordinates of the pearabela vertices, the coordinates of the point of intersection of the graphics of the function with the coordinate axes, according to the function schedule to determine the properties of the quadratic function.

Equipment:

Lesson plan

Organizational moment (1-2 min)

Actual knowledge of knowledge (10 min)

Statement of new material (15 min)

Fixing a new material (12 min)

Summing up (3 min)

Home Task (2 min)

During the classes

Organizing time

Greeting, checking absent, collecting notebooks.

Actualization of knowledge

Teacher: In today's lesson, we will study the new topic: "Function". But to begin with, we repeat the previously studied material.

Frontal survey:

What is called a quadratic function? (Function where specified valid numbers, valid variable, is called a quadratic function.)

What is a chart of a quadratic function? (A chart of a quadratic function is parabola.)

What is zeros of a quadratic function? (Zeros of the quadratic function - the values \u200b\u200bat which it turns into zero.)

List the properties of the function. (The values \u200b\u200bof the function are positive with and equal to zero at; the graph of the function is symmetrical with respect to the ordinate OS; when the function increases, when - decreases.)

List the properties of the function. (If, the function takes positive values \u200b\u200bwhen, if the function takes negative values \u200b\u200bwhen, the value of the function is 0 only; Parabola is symmetrical with respect to the axis of the ordinate; if, the function increases with and decreases when, then the function increases, decreases. at.)

Statement of new material

Teacher: Let's start learning a new material. Open the notebook, write down the number and theme of the lesson. Pay attention to the board.

Recording on the board: number.

Function.

Teacher: On the board you see two graphics of functions. First chart, and the second. Let's try to compare them.

The properties of the function you know. On their basis, and comparing our graphs, you can select the properties of the function.

So, what do you think, what will the direction of the branches of the parabola will depend on?

Pupils: The direction of the branches of both parabola will depend on the coefficient.

Teacher: Completely right. You can also see that both parabolas have a symmetry axis. At the first schedule function, what is the axis of symmetry?

Pupils: Parabola is a type of symmetry axis is the ordinate axis.

Teacher: True. And what is the axis of symmetry parabola

Pupils: The axis of symmetry Parabola is a line that passes through the top of the parabola, parallel to the axis of the ordinate.

Teacher: That's right. So, the axis of the symmetry of the graph of the function will be called a direct, passing through the top of the parabola, parallel axis of the ordinate.

And the top of the parabola is a point with coordinates. They are determined by the formula:

Write down the formula in the notebook and circle into the frame.

Recording on the board and in notebooks

The coordinates of the pearabol vertices.

Teacher: Now, to be more clear, consider an example.

Example 1: Find the coordinates of the pearabela vertices .

Solution: by formula

Teacher: As we have noted, the axis of symmetry passes through the peak of the parabola. Look at the desk. Distribute this drawing in the notebook.

Recording on the board and in notebooks:

Teacher: In the drawing: - The equation of the axis of the symmetry of the parabola with the vertex at the point where the abscissa of the pearabol vertices.

Consider an example.

Example 2: According to the graph of the function, determine the equation of the axis of the parabola symmetry.

The equation of the axis of symmetry has the form:, therefore, the equation of the symmetry axis of this parabola.

Answer: - equation of the axis of symmetry.

Fastening a new material

Teacher: On the board recorded tasks that need to be solved in the classroom.

Recording on the board: No. 609 (3), 612 (1), 613 (3)

Teacher: But at first I decided an example from the textbook. We will decide at the board.

Example 1: Find the coordinates of the vertex parabola

Solution: by formula

Answer: The coordinates of the pearabol vertex.

Example 2: Find the coordinates of the points of intersection of parabola with coordinate axes.

Solution: 1) with the axis:

Those.

On the Vieta Theorem:

The intersection points with the abscissa axis (1; 0) and (2; 0).

Presentation and lesson on the topic:
"Function schedule $ Y \u003d AX ^ 2 + BX + C $. Properties"

Additional materials
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Training manuals and simulators in the online store "Integral" for grade 8
Manual for the textbook Dorofeeva G.V. Manual for the textbook of Nikolsky S.M.

Guys, we built on the last lessons a large number of Graphs, including a lot of parabola. Today we summarize the knowledge gained and learn how to build graphs of this function in the most general form.
Let's look at the square three of $ a * x ^ 2 + b * x + C $. $ a, b, C $ are called coefficients. They can be any numbers, but $ a ≠ $ 0. $ A * X ^ 2 $ is called a senior member, $ a $ is a senior coefficient. It is worth noting that the coefficients of $ b $ and $ C $ can be zero, that is, three decreases will consist of two members, and the third is zero.

Let's look at the function $ y \u003d a * x ^ 2 + b * x + C $. This feature is called "quadratic" because elder degree Second, that is, the square. The coefficients are the same as defined above.

At the last lesson in the last example, we disassemble the construction of a graph of a similar function.
Let's prove that any such quadratic function can be reduced to mind: $ y \u003d a (x + l) ^ 2 + m $.

The schedule of such a function is built using an additional coordinate system. In a large mathematics, the numbers are quite rare. Virtually any task is required to prove in the general case. Today we will analyze one of these evidence. Guys, you can, see the power of the mathematical apparatus, but also its complexity.

We highlight the full square of square three declections:
$ a * x ^ 2 + b * x + c \u003d (a * x ^ 2 + b * x) + c \u003d a (x ^ 2 + \\ FRAC (B) (a) * x) + C \u003d $ $ \u003d A (X ^ 2 + 2 \\ FRAC (B) (2a) * x + \\ FRAC (B ^ 2) (4a)) - \\ FRAC (B ^ 2) (4A) + C \u003d A (X + \\ FRAC (B) (2a)) ^ 2+ \\ FRAC (4AC-B ^ 2) (4a) $.
We got what they wanted.
Any quadratic function can be represented as:
$ y \u003d a (x + l) ^ 2 + m $, where $ L \u003d \\ FRAC (B) (2a) $, $ M \u003d \\ FRAC (4AC-B ^ 2) (4a) $.

To build a graph $ y \u003d a (x + L) ^ 2 + m $, you need to build a graph of the function $ y \u003d ax ^ 2 $. And the top of the parabola will be at the point with the coordinates of $ (- l; m) $.
So, our function is $ y \u003d a * x ^ 2 + b * x + C $ - parabol.
The parabola axis will be straight $ x \u003d - \\ Frac (b) (2a) $, and the coordinates of the pearabol vertex along the abscissa axis, as we can notice, is calculated by the formula: $ x_ (B) \u003d - \\ FRAC (B) (2a) $.
To calculate the coordinates of the vertex parabola along the ordinate axis, you can:

• use the formula: $ y_ (c) \u003d \\ FRAC (4AC-B ^ 2) (4a) $
• directly substitute in the initial function of the coordinate of the vertex of $ x $: $ y_ (B) \u003d AX_ (B) ^ 2 + b * x_ (B) + C $.

How to calculate the ordinate of the vertex? Again, the choice is yours, but usually it will be easier to consider the second way.
If you want to describe some properties or respond to some specific questions, you do not always need to build a function schedule. The main questions that can be answered without building, consider in the following example.

Example 1.
Without constructing the schedule of the function $ y \u003d 4x ^ 2-6x-3 $, answer the following questions:

Decision.
a) the parabol axis is direct $ X \u003d - \\ FRAC (B) (2a) \u003d - \\ FRAC (-6) (2 * 4) \u003d \\ FRAC (6) (8) \u003d \\ FRAC (3) (4) $ .
b) The abscissa of the vertices we found above $ x_ (B) \u003d \\ FRAC (3) (4) $.
The ordinate of the vertices will find the direct substitution in the original function:
$ y_ (c) \u003d 4 * (\\ FRAC (3) (4)) ^ 2-6 * \\ FRAC (3) (4) -3 \u003d \\ FRAC (9) (4) - \\ FRAC (18) (4 ) - \\ FRAC (12) (4) \u003d - \\ FRAC (21) (4) $.
c) the graph required by the function will be parallel to the transfer of the schedule $ y \u003d 4x ^ 2 $. Its branches are looking up, and therefore branches of parabolas of the original function will also look up.
In general, if the coefficient is $ a\u003e $ 0, then the branches are watching up if $ a coefficient
Example 2.
Build a function graph: $ y \u003d 2x ^ 2 + 4x-6 $.

Decision.
We will find the coordinates of the vertices of Parabola:
$ x_ (B) \u003d - \\ FRAC (B) (2a) \u003d - \\ FRAC (4) (4) \u003d - 1 $.
$ y_ (B) \u003d 2 * (- 1) ^ 2 + 4 (-1) -6 \u003d 2-4-6 \u003d -8 $.
We note the coordinate of the vertex on the coordinate axis. At this point, as if in the new coordinate system, we construct a parabol $ y \u003d 2x ^ 2 $.

There are many ways to simplify the construction of parabola charts.

• We can find two symmetric points, calculate the value of the function at these points, mark them on the coordinate plane and connect them to the vertex curve describing the parabola.
• We can build a parabola branch to the right or the left of the top and then reflect it.
• We can build by points.

Example 3.
Find the highest I. the smallest value Functions: $ y \u003d -x ^ 2 + 6x + 4 $ on a segment $ [- 1; 6] $.

Decision.
We build a graph of this feature, select the required gap and find the lowest and highest point of our schedule.
We will find the coordinates of the vertices of Parabola:
$ x_ (B) \u003d - \\ FRAC (B) (2a) \u003d - \\ FRAC (6) (- 2) \u003d $ 3.
$ y_ (c) \u003d - 1 * (3) ^ 2 + 6 * 3 + 4 \u003d -9 + 18 + 4 \u003d 13 $.
At the point with coordinates $ (3; 13) $ we construct a parabol $ y \u003d -x ^ 2 $. Select the required gap. The lowest point has coordinate -3, the highest point - coordinate 13.
$ y_ (nim) \u003d - $ 3; $ y_ (naib) \u003d 13 $.

Tasks for self solutions

1. Without constructing a schedule of the function $ Y \u003d -3x ^ 2 + 12x-4 $, answer the following questions:
a) Specify a straight line that serves the axis of parabola.
B) Find the vertices coordinates.
c) Where does parabola look (up or down)?
2. Build a function graph: $ y \u003d 2x ^ 2-6x + $ 2.
3. Build a graph of the function: $ Y \u003d -X ^ 2 + 8x-4 $.
4. Find the most and the smallest function of the function: $ y \u003d x ^ 2 + 4x-3 $ on the segment $ [- 5; 2] $.

Lesson: How to build a parabola or quadratic function?

THEORETICAL PART

Parabola is a graph of the function described by formula AX 2 + BX + C \u003d 0.
To build a parabola need to follow a simple action algorithm:

1) Parabola formula Y \u003d AX 2 + BX + C,
if a a\u003e 0. then the branches of parabola are directed up,
and the branches of parabola are directed down.
Free dick c. This point crosses the parabola with the Oy axis;

2), it is found according to the formula x \u003d (- b) / 2a, found x we \u200b\u200bsubstitute in the parabola equation and find y.;

3) Zero function Or, on another point of intersection of the parabola with the OX axis, they are also called the roots of the equation. To find the roots we equate equate to 0 aX 2 + BX + C \u003d 0;

Types of equations:

a) Full quadratic equation Has appearance AX 2 + BX + C \u003d 0and is solved by discriminant;
b) incomplete square equation AX 2 + BX \u003d 0. To solve it, you need to make x for brackets, then each multiplier to equate to 0:
AX 2 + BX \u003d 0,
x (ax + b) \u003d 0,
x \u003d 0 and ax + b \u003d 0;
c) incomplete square equation AX 2 + C \u003d 0. To solve it, unknown to transfer one way, and known to another. x \u003d ± √ (C / A);

4) Find a few additional points to build a function.

Practical part

And so now on the example we will analyze all the actions:
Example number 1:
y \u003d x 2 + 4x + 3
C \u003d 3 means Parabola crosses OY at point x \u003d 0 y \u003d 3. Parabola branches look up as a \u003d 1 1\u003e 0.
a \u003d 1 b \u003d 4 c \u003d 3 x \u003d (- b) / 2a \u003d (- 4) / (2 * 1) \u003d - 2 y \u003d (-2) 2 +4 * (- 2) + 3 \u003d 4- 8 + 3 \u003d -1 Top is at point (-2; -1)
Find the roots of the equation x 2 + 4x + 3 \u003d 0
On discriminative find roots
a \u003d 1 b \u003d 4 c \u003d 3
D \u003d b 2 -4ac \u003d 16-12 \u003d 4
x \u003d (- b ± √ (d)) / 2a
x 1 \u003d (- 4 + 2) / 2 \u003d -1
x 2 \u003d (- 4-2) / 2 \u003d -3

Take several arbitrary points that are near the top x \u003d -2

x -4 -3 -1 0
3 0 0 3

We substitute instead of x in the equation y \u003d x 2 + 4x + 3 values
y \u003d (- 4) 2 +4 * (- 4) + 3 \u003d 16-16 + 3 \u003d 3
Y \u003d (- 3) 2 +4 * (- 3) + 3 \u003d 9-12 + 3 \u003d 0
y \u003d (- 1) 2 +4 * (- 1) + 3 \u003d 1-4 + 3 \u003d 0
y \u003d (0) 2 + 4 * (0) + 3 \u003d 0-0 + 3 \u003d 3
Seen by the values \u200b\u200bof the function that the parabol is symmetric with respect to direct x \u003d -2

Example number 2:
y \u003d -X 2 + 4X
c \u003d 0 So Parabola crosses OY at point x \u003d 0 y \u003d 0. Parabola branches look down as a \u003d -1 -1 Find the roots of the equation -x 2 + 4x \u003d 0
An incomplete square equation of AX 2 + BX \u003d 0. To decide it, you need to make x for brackets, then each multiplier to equate to 0.
x (-x + 4) \u003d 0, x \u003d 0 and x \u003d 4.

Take several arbitrary points that are near the top x \u003d 2
x 0 1 3 4
0 3 3 0
We substitute instead of the equation y \u003d -x 2 + 4x values
Y \u003d 0 2 + 4 * 0 \u003d 0
y \u003d - (1) 2 + 4 * 1 \u003d -1 + 4 \u003d 3
Y \u003d - (3) 2 + 4 * 3 \u003d -9 + 13 \u003d 3
Y \u003d - (4) 2 + 4 * 4 \u003d -16 + 16 \u003d 0
It can be seen by the values \u200b\u200bof the function that Parabola is symmetric with respect to direct x \u003d 2

Example number 3.
y \u003d x 2 -4
C \u003d 4 So Parabola crosses OY at point x \u003d 0 y \u003d 4. Parabola branches look up as a \u003d 1 1\u003e 0.
a \u003d 1 b \u003d 0 C \u003d -4 x \u003d (- b) / 2a \u003d 0 / (2 * (1)) \u003d 0 y \u003d (0) 2 -4 \u003d -4 vertex is at point (0; -4 )
Find the roots of the equation x 2 -4 \u003d 0
An incomplete square equation of the form AX 2 + C \u003d 0. To solve it, unknown to transfer one way, and known to another. x \u003d ± √ (C / A)
x 2 \u003d 4
x 1 \u003d 2
x 2 \u003d -2

Take a few arbitrary points that are near the top x \u003d 0
x -2 -1 1 2
0 -3 -3 0
We substitute instead of x equation y \u003d x 2 -4 values
Y \u003d (- 2) 2 -4 \u003d 4-4 \u003d 0
Y \u003d (- 1) 2 -4 \u003d 1-4 \u003d -3
Y \u003d 1 2 -4 \u003d 1-4 \u003d -3
y \u003d 2 2 -4 \u003d 4-4 \u003d 0
Seen by the values \u200b\u200bof the function that parabola is symmetrical with respect to direct x \u003d 0

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Tasks for properties and graphics quadratic function Call, as practice shows, serious difficulties. It is rather strange, because the quadratic function is held in the 8th grade, and then the entire first quarter of the 9th grade "survive" the properties of the parabola and build its graphs for various parameters.

This is due to the fact that forcing students to build parabolas, almost do not pay time for reading charts, that is, not practicing the understanding of the information obtained from the picture. Apparently, it is assumed that by building a dozen two charts, a smart schoolboy will detect himself and formulates the connection of coefficients in the formula and appearance graphics. In practice it does not work. For such a generalization, a serious experience of mathematical mini studies, which most nine-graduates, of course, do not have it. Meanwhile, in GIA suggest precisely on the schedule to determine the signs of coefficients.

Let's not require schoolchildren impossible and simply offer one of the algorithms to solve such problems.

So, the function of the form y \u003d AX 2 + BX + C It is called a quadratic, the schedule is parabola. As follows from the name, the main term is aX 2.. I.e but should not be zero, the remaining coefficients ( b. and from) can be zero.

Let's see how the signs of its coefficients affect the appearance of the parabola.

The simplest dependence for the coefficient but. Most schoolchildren confidently replies: "If but \u003e 0, then the parabola branches are directed upwards, and if but < 0, - то вниз". Совершенно верно. Ниже приведен график квадратичной функции, у которой but > 0.

y \u003d 0.5X 2 - 3X + 1

In this case but = 0,5

And now for but < 0:

y \u003d - 0.5x2 - 3x + 1

In this case but = - 0,5

Influence of the coefficient from Also easy to trace enough. Imagine that we want to find the value of the function at the point h. \u003d 0. Substitute Zero in the formula:

y. = a. 0 2 + b. 0 + c. = c.. Turns out that y \u003d s. I.e from - This is the ordinate of the point of intersection of the parabola with the axis. As a rule, this point is easy to find on the chart. And determine above zero it lies or below. I.e from \u003e 0 or from < 0.

from > 0:

y \u003d x 2 + 4x + 3

from < 0

y \u003d x 2 + 4x - 3

Accordingly, if from \u003d 0, then Parabola will definitely pass through the origin of the coordinate:

y \u003d x 2 + 4x

More difficult with the parameter b.. The point on which we will find it depends not only from b. But from but. This is the top of the parabola. Its abscissa (axis coordinate h.) is on the formula x B \u003d - b / (2a). In this way, b \u003d - 2ach in. That is, we act as follows: on the chart we find the top of the parabola, we define the sign of its abscissa, that is, we look to the right of zero ( x B. \u003e 0) or left ( x B. < 0) она лежит.

However, this is not all. We also need to pay attention to the coefficient sign but. That is, to see where the branches of parabola are directed. And only after that by the formula b \u003d - 2ach in Determine the sign b..

Consider an example:

Branches are directed up, it means but \u003e 0, Parabola crosses the axis w. below zero, then from < 0, вершина параболы лежит правее нуля. Следовательно, x B. \u003e 0. So b \u003d - 2ach in = -++ = -. b. < 0. Окончательно имеем: but > 0, b. < 0, from < 0.